In ΔABC ∠B = 90° [∵ Given] Therefore, ∠A < 90° and ∠C ∠C [∵ ∠B = 90° and ∠C AB ...(1) [∵ In a triangle, greater angle has longer side opposite to it] Similarly, in Δ ABC, ∠B > ∠A [∵ ∠B = 90° and ∠A BC ...(2) [∵ In a triangle, greater angle has longer side opposite to it] From the equation (1)Read more
In ΔABC ∠B = 90° [∵ Given]
Therefore, ∠A < 90° and ∠C ∠C [∵ ∠B = 90° and ∠C AB …(1) [∵ In a triangle, greater angle has longer side opposite to it]
Similarly, in Δ ABC, ∠B > ∠A [∵ ∠B = 90° and ∠A BC …(2) [∵ In a triangle, greater angle has longer side opposite to it]
From the equation (1) and (2), we have
AC > AB and AC > BC
Hence, hypotenuse AC is the longest side.
Construction: join AC. In ΔABC, BC > AB [∵ AB is the shortest side of the quadrilateral ABCD] Hence, ∠1 > ∠3 ...(1) [∵ In a triangle, longer side has greater angle opposite to it] Similarly, In ΔADC, CD > AD [∵ CD is the longest side of the quadrilateral ABCD] Hence, ∠2>∠3 + ∠4 ...(2) [∵Read more
Construction: join AC.
In ΔABC, BC > AB [∵ AB is the shortest side of the quadrilateral ABCD]
Hence, ∠1 > ∠3 …(1) [∵ In a triangle, longer side has greater angle opposite to it]
Similarly,
In ΔADC,
CD > AD [∵ CD is the longest side of the quadrilateral ABCD]
Hence, ∠2>∠3 + ∠4 …(2) [∵ In a triangle, Longer side has greater angle opposite to it]
From the equation (1) and (2), we have
∠1 +∠2 > ∠3 + ∠4
⇒ ∠A> ∠C
Construction: Join BD.
In ABD, AD> AB [∵ AB is the shortest side of the quadrilateral ABCD]
Hence, ∠5 > ∠7 …(3) [∵ In a triangle, longer side has greater angle opposite to it]
Similarly.
In ΔBDC, CD > BC [∵ CD is the longest side of the quadrilateral ABCD]
Hence, ∠6 > ∠8 …(4) [∵ In a triangle, longer side has greater angle opposite to it]
From the equation (3) and (4), we have
∠5 + ∠6 > ∠7 + ∠8 ⇒ ∠B > ∠D.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
ABC is an isosceles triangle with AB = AC. Draw AP⊥BC to show that Angle B = Angle C.
Show that in a right angled triangle, the hypotenuse is the longest side.
In ΔABC ∠B = 90° [∵ Given] Therefore, ∠A < 90° and ∠C ∠C [∵ ∠B = 90° and ∠C AB ...(1) [∵ In a triangle, greater angle has longer side opposite to it] Similarly, in Δ ABC, ∠B > ∠A [∵ ∠B = 90° and ∠A BC ...(2) [∵ In a triangle, greater angle has longer side opposite to it] From the equation (1)Read more
In ΔABC ∠B = 90° [∵ Given]
See lessTherefore, ∠A < 90° and ∠C ∠C [∵ ∠B = 90° and ∠C AB …(1) [∵ In a triangle, greater angle has longer side opposite to it]
Similarly, in Δ ABC, ∠B > ∠A [∵ ∠B = 90° and ∠A BC …(2) [∵ In a triangle, greater angle has longer side opposite to it]
From the equation (1) and (2), we have
AC > AB and AC > BC
Hence, hypotenuse AC is the longest side.
In Figure sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠ PBC AB.
∠PBC + ∠ABC = 180° [∵ Linear Pair] ⇒ ∠PBC = 180° - ∠ABC ...(1) Similarly, ∠QCB + ∠ACB = 180° [∵ Linear pair] ⇒∠QCB = 180° - ∠ACB ...(2) Given that: ∠PBC > ∠QCB ⇒ 180° - ∠ABC - ∠ACB ⇒ ∠ABC > ∠ACB ...(3) In ΔABC, ∠ABC > ∠ACB [∵ From the equation (3)] Hence, AC > AB [∵ In a triangle, greateRead more
∠PBC + ∠ABC = 180° [∵ Linear Pair]
See less⇒ ∠PBC = 180° – ∠ABC …(1)
Similarly,
∠QCB + ∠ACB = 180° [∵ Linear pair]
⇒∠QCB = 180° – ∠ACB …(2)
Given that:
∠PBC > ∠QCB
⇒ 180° – ∠ABC – ∠ACB
⇒ ∠ABC > ∠ACB …(3)
In ΔABC,
∠ABC > ∠ACB [∵ From the equation (3)]
Hence, AC > AB [∵ In a triangle, greater angle has longer side opposite to it]
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Figure). Show that ∠ A > ∠ C and∠ B > ∠ D.
Construction: join AC. In ΔABC, BC > AB [∵ AB is the shortest side of the quadrilateral ABCD] Hence, ∠1 > ∠3 ...(1) [∵ In a triangle, longer side has greater angle opposite to it] Similarly, In ΔADC, CD > AD [∵ CD is the longest side of the quadrilateral ABCD] Hence, ∠2>∠3 + ∠4 ...(2) [∵Read more
Construction: join AC.
See lessIn ΔABC, BC > AB [∵ AB is the shortest side of the quadrilateral ABCD]
Hence, ∠1 > ∠3 …(1) [∵ In a triangle, longer side has greater angle opposite to it]
Similarly,
In ΔADC,
CD > AD [∵ CD is the longest side of the quadrilateral ABCD]
Hence, ∠2>∠3 + ∠4 …(2) [∵ In a triangle, Longer side has greater angle opposite to it]
From the equation (1) and (2), we have
∠1 +∠2 > ∠3 + ∠4
⇒ ∠A> ∠C
Construction: Join BD.
In ABD, AD> AB [∵ AB is the shortest side of the quadrilateral ABCD]
Hence, ∠5 > ∠7 …(3) [∵ In a triangle, longer side has greater angle opposite to it]
Similarly.
In ΔBDC, CD > BC [∵ CD is the longest side of the quadrilateral ABCD]
Hence, ∠6 > ∠8 …(4) [∵ In a triangle, longer side has greater angle opposite to it]
From the equation (3) and (4), we have
∠5 + ∠6 > ∠7 + ∠8 ⇒ ∠B > ∠D.