(i) In ΔABD and ΔACD AB = AC [∵ Given] BD = CD [∵ Given] AD = AD [∵ Common] Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule] (ii) In ΔABD ≅ ΔACD [∵ Proved above] ∠ BAD = ∠ CAD [∵ CPCT] In ABP and ACP, AB = AC [∵ Given] ∠ BAP = ∠ CAP [∵ Proved above] AP = AP [∵ Common] Hence, ΔABP ≅ ΔACP [∵ SAS CongruencyRead more
(i) In ΔABD and ΔACD
AB = AC [∵ Given]
BD = CD [∵ Given]
AD = AD [∵ Common]
Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule]
(ii) In ΔABD ≅ ΔACD [∵ Proved above]
∠ BAD = ∠ CAD [∵ CPCT]
In ABP and ACP,
AB = AC [∵ Given]
∠ BAP = ∠ CAP [∵ Proved above]
AP = AP [∵ Common]
Hence, ΔABP ≅ ΔACP [∵ SAS Congruency rule]
(iii) In ΔABD ≅ ΔACD [∵ Proved above]
∠ BAD = ∠ CAD [∵ CPCT]
∠ BDA = ∠ CDA [∵ CPCT]
Hence, AP bisects both the angles A and D
(iv) In ΔABP ≅ ΔACP [∵ Prove above]
BP = CP [∵ CPCT]
∠BPA = ∠CPA [∵ CPCT]
∠BPA + ∠CPA = 180° [∵ Linear Pair]
⇒ ∠CPA + ∠CPA = 180° [∵ ∠BPA = ∠CPA]
⇒ 2∠ CPA = 180° ⇒ ∠CPA = (180°/2) = 90°
⇒ AP is perpendicular to BC. ⇒ AP is perpendicular bisector of BC. [∵ BP = CP].
In ΔABD, E is mid-point of AD [∵ Given] and EG ∥ AB [∵ Given] Hence, G is mid-point of BD [∵ Converse of Mid-point Theorem] Similarly, In ΔBCD, G is mid-point of BD [∵ Prove above] and FG ∥ DC [∵ Given] Hence, F is mid-point of BC [∵ Converse of Mid-point Theorem]
In ΔABD,
E is mid-point of AD [∵ Given]
and EG ∥ AB [∵ Given]
Hence, G is mid-point of BD [∵ Converse of Mid-point Theorem]
Similarly,
In ΔBCD,
G is mid-point of BD [∵ Prove above]
and FG ∥ DC [∵ Given]
Hence, F is mid-point of BC [∵ Converse of Mid-point Theorem]
In AECD, AE ∥ DC [∵ Given] AD ∥ CE [∵ By construction] Hence, AECD is a parallelogram. AD = CE ...(1) [∵ Opposite sides of a parallelogram are equal] AD = BC ...(2) [∵ Given] Hence, CE = BC [∵ From the equation (1) and (2)] Therefore, in ΔBCE, ∠3 = ∠4 ...(3) [∵ In a triangle, the angles opposite toRead more
In AECD,
AE ∥ DC [∵ Given]
AD ∥ CE [∵ By construction]
Hence, AECD is a parallelogram.
AD = CE …(1) [∵ Opposite sides of a parallelogram are equal]
AD = BC …(2) [∵ Given]
Hence, CE = BC [∵ From the equation (1) and (2)]
Therefore, in ΔBCE,
∠3 = ∠4 …(3) [∵ In a triangle, the angles opposite to equal sides are equal]
Here, ∠2 + ∠3 = 180° …(4) [∵ Linear Pair]
∠1 + ∠4 = 180° …(5) [∵ Co-interior angles]
Therefore, ∠2 + ∠3 = ∠1 + ∠4 [∵ From the equation (4) and (5)]
⇒ ∠2 = ∠1 ⇒ ∠B = ∠A [∵ ∠3 = ∠4]
(ii) ABCD ia a trapezium in which AB ∥ DC, hence,
∠1 + ∠D = 180° …(6) [∵ Co-interior angles]
∠2 + ∠C = 180° …(7) [∵ Co-interior angles]
Therefore, ∠1 + ∠D = ∠2 + C [∵ From the equation (6) and (7)]
⇒ ∠D = ∠C [∵ ∠2 = ∠1]
(iii) In ΔABC and ΔBAD,
BC = AD [∵ Given]
∠ABC = ∠BAD [∵ Prove above]
AB = AB [∵ Common]
Hence, ΔABC ≅ ΔBAD [∵ SAS Congruency rule]
(iv) ΔABC ≅ ΔBAD [∵ Prove above]
Diagonal AC = diagonal BD [∵ CPCT]
Consider a circle with centre O. Let AB is the given line. Now draw a perpendicular from O to line AB, which intersect AB at P. Now take points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.
Consider a circle with centre O. Let AB is the given line.
Now draw a perpendicular from O to line AB, which intersect AB at P.
Now take points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.
Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Figure). If AD is extended to intersect BC at P, show that (i) Δ ABD ≅ Δ ACD (ii) Δ ABP ≅ Δ ACP (iii) AP bisects ∠ A as well as ∠ D. (iv) AP is the perpendicular bisector of BC.
(i) In ΔABD and ΔACD AB = AC [∵ Given] BD = CD [∵ Given] AD = AD [∵ Common] Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule] (ii) In ΔABD ≅ ΔACD [∵ Proved above] ∠ BAD = ∠ CAD [∵ CPCT] In ABP and ACP, AB = AC [∵ Given] ∠ BAP = ∠ CAP [∵ Proved above] AP = AP [∵ Common] Hence, ΔABP ≅ ΔACP [∵ SAS CongruencyRead more
(i) In ΔABD and ΔACD
See lessAB = AC [∵ Given]
BD = CD [∵ Given]
AD = AD [∵ Common]
Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule]
(ii) In ΔABD ≅ ΔACD [∵ Proved above]
∠ BAD = ∠ CAD [∵ CPCT]
In ABP and ACP,
AB = AC [∵ Given]
∠ BAP = ∠ CAP [∵ Proved above]
AP = AP [∵ Common]
Hence, ΔABP ≅ ΔACP [∵ SAS Congruency rule]
(iii) In ΔABD ≅ ΔACD [∵ Proved above]
∠ BAD = ∠ CAD [∵ CPCT]
∠ BDA = ∠ CDA [∵ CPCT]
Hence, AP bisects both the angles A and D
(iv) In ΔABP ≅ ΔACP [∵ Prove above]
BP = CP [∵ CPCT]
∠BPA = ∠CPA [∵ CPCT]
∠BPA + ∠CPA = 180° [∵ Linear Pair]
⇒ ∠CPA + ∠CPA = 180° [∵ ∠BPA = ∠CPA]
⇒ 2∠ CPA = 180° ⇒ ∠CPA = (180°/2) = 90°
⇒ AP is perpendicular to BC. ⇒ AP is perpendicular bisector of BC. [∵ BP = CP].
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F see figure. Show that F is the mid-point of BC.
In ΔABD, E is mid-point of AD [∵ Given] and EG ∥ AB [∵ Given] Hence, G is mid-point of BD [∵ Converse of Mid-point Theorem] Similarly, In ΔBCD, G is mid-point of BD [∵ Prove above] and FG ∥ DC [∵ Given] Hence, F is mid-point of BC [∵ Converse of Mid-point Theorem]
In ΔABD,
See lessE is mid-point of AD [∵ Given]
and EG ∥ AB [∵ Given]
Hence, G is mid-point of BD [∵ Converse of Mid-point Theorem]
Similarly,
In ΔBCD,
G is mid-point of BD [∵ Prove above]
and FG ∥ DC [∵ Given]
Hence, F is mid-point of BC [∵ Converse of Mid-point Theorem]
ABCD is a trapezium in which AB || CD and AD = BC see Figure. Show that (i) ∠ A = ∠ B (ii) ∠ C = ∠ D (iii) ∆ ABC ≅ ∆ BAD (iv) diagonal AC = diagonal BD
In AECD, AE ∥ DC [∵ Given] AD ∥ CE [∵ By construction] Hence, AECD is a parallelogram. AD = CE ...(1) [∵ Opposite sides of a parallelogram are equal] AD = BC ...(2) [∵ Given] Hence, CE = BC [∵ From the equation (1) and (2)] Therefore, in ΔBCE, ∠3 = ∠4 ...(3) [∵ In a triangle, the angles opposite toRead more
In AECD,
See lessAE ∥ DC [∵ Given]
AD ∥ CE [∵ By construction]
Hence, AECD is a parallelogram.
AD = CE …(1) [∵ Opposite sides of a parallelogram are equal]
AD = BC …(2) [∵ Given]
Hence, CE = BC [∵ From the equation (1) and (2)]
Therefore, in ΔBCE,
∠3 = ∠4 …(3) [∵ In a triangle, the angles opposite to equal sides are equal]
Here, ∠2 + ∠3 = 180° …(4) [∵ Linear Pair]
∠1 + ∠4 = 180° …(5) [∵ Co-interior angles]
Therefore, ∠2 + ∠3 = ∠1 + ∠4 [∵ From the equation (4) and (5)]
⇒ ∠2 = ∠1 ⇒ ∠B = ∠A [∵ ∠3 = ∠4]
(ii) ABCD ia a trapezium in which AB ∥ DC, hence,
∠1 + ∠D = 180° …(6) [∵ Co-interior angles]
∠2 + ∠C = 180° …(7) [∵ Co-interior angles]
Therefore, ∠1 + ∠D = ∠2 + C [∵ From the equation (6) and (7)]
⇒ ∠D = ∠C [∵ ∠2 = ∠1]
(iii) In ΔABC and ΔBAD,
BC = AD [∵ Given]
∠ABC = ∠BAD [∵ Prove above]
AB = AB [∵ Common]
Hence, ΔABC ≅ ΔBAD [∵ SAS Congruency rule]
(iv) ΔABC ≅ ΔBAD [∵ Prove above]
Diagonal AC = diagonal BD [∵ CPCT]
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 12 cm. Length PQ is :
In ΔOPQ, angle P is right angle. [Since radius is perpendicular to tangent] Using Pythagoras theorem, OQ² = PQ² + OP² ⇒ 12² = PQ² + 5² ⇒ 144 = PQ² + 25 ⇒ PQ² = 144 - 25 = 119 ⇒ PQ = √119 Hence, the option (D) is correct.
In ΔOPQ, angle P is right angle. [Since radius is perpendicular to tangent]
See lessUsing Pythagoras theorem, OQ² = PQ² + OP²
⇒ 12² = PQ² + 5²
⇒ 144 = PQ² + 25
⇒ PQ² = 144 – 25 = 119
⇒ PQ = √119
Hence, the option (D) is correct.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Consider a circle with centre O. Let AB is the given line. Now draw a perpendicular from O to line AB, which intersect AB at P. Now take points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.
Consider a circle with centre O. Let AB is the given line.
See lessNow draw a perpendicular from O to line AB, which intersect AB at P.
Now take points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.