In ΔABD and ΔACD, BD = DC [∵ AD bisects BC] ∠ADB = ∠ADC [∵ Each 90°] AD = AD [∵ Common] Hance, ΔABD ≅ ΔACD [∵ SAS Congruency Rule] AB = AC [∵ CPCT] Hence, ABC is an isosceles triangle.
In ΔABD and ΔACD,
BD = DC [∵ AD bisects BC]
∠ADB = ∠ADC [∵ Each 90°]
AD = AD [∵ Common]
Hance, ΔABD ≅ ΔACD [∵ SAS Congruency Rule]
AB = AC [∵ CPCT]
Hence, ABC is an isosceles triangle.
(i) In ΔABE and ΔACF, ∠AEB = ∠AFC [∵ Each 90] ∠A = ∠A [∵ Common] BE = CF [∵ Given] Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule] (ii) In ΔABE ΔACF [∵ Proved above] AB = AC [∵ CPCT] Hence, ΔABC is an isosceles triangle.
(i) In ΔABE and ΔACF,
∠AEB = ∠AFC [∵ Each 90]
∠A = ∠A [∵ Common]
BE = CF [∵ Given]
Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule]
(ii) In ΔABE ΔACF [∵ Proved above]
AB = AC [∵ CPCT]
Hence, ΔABC is an isosceles triangle.
In ΔABC, AB = AC [∵ Given] ∠ABC = ∠ACB ...(1) [∵ Angles opposite to equal sides are equal] In ΔDBC, DB = DC [∵ Given] ∠DBC = ∠DCB ...(2) [∵ Angles opposite to equal sides are equal] Adding equation (1) and (2), we get ∠ABC + ∠DBC = ∠ACB + ∠DCB ⇒ ∠ABD = ∠ACD.
In ΔABC,
AB = AC [∵ Given]
∠ABC = ∠ACB …(1) [∵ Angles opposite to equal sides are equal]
In ΔDBC,
DB = DC [∵ Given]
∠DBC = ∠DCB …(2) [∵ Angles opposite to equal sides are equal]
Adding equation (1) and (2), we get
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒ ∠ABD = ∠ACD.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR (see Figure). Show that: (i) Δ ABM ≅ Δ PQN (ii) Δ ABC ≅ Δ PQR
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Figure). Show that these altitudes are equal.
In ΔABE and ΔACF, ∠AEB = ∠AFC [∵ Each 90°] ∠A = ∠A [∵ Common] AB = AC [∵ Given] Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule] BE = CF [∵ CPCT]
In ΔABE and ΔACF,
See less∠AEB = ∠AFC [∵ Each 90°]
∠A = ∠A [∵ Common]
AB = AC [∵ Given]
Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule]
BE = CF [∵ CPCT]
In Δ ABC, AD is the perpendicular bisector of BC (see Figure). Show that Δ ABC is an isosceles triangle in which AB = AC.
In ΔABD and ΔACD, BD = DC [∵ AD bisects BC] ∠ADB = ∠ADC [∵ Each 90°] AD = AD [∵ Common] Hance, ΔABD ≅ ΔACD [∵ SAS Congruency Rule] AB = AC [∵ CPCT] Hence, ABC is an isosceles triangle.
In ΔABD and ΔACD,
See lessBD = DC [∵ AD bisects BC]
∠ADB = ∠ADC [∵ Each 90°]
AD = AD [∵ Common]
Hance, ΔABD ≅ ΔACD [∵ SAS Congruency Rule]
AB = AC [∵ CPCT]
Hence, ABC is an isosceles triangle.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Figure). Show that (i) Δ ABE ≅ Δ ACF (ii) AB = AC, i.e., ABC is an isosceles triangle.
(i) In ΔABE and ΔACF, ∠AEB = ∠AFC [∵ Each 90] ∠A = ∠A [∵ Common] BE = CF [∵ Given] Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule] (ii) In ΔABE ΔACF [∵ Proved above] AB = AC [∵ CPCT] Hence, ΔABC is an isosceles triangle.
(i) In ΔABE and ΔACF,
∠AEB = ∠AFC [∵ Each 90]
∠A = ∠A [∵ Common]
BE = CF [∵ Given]
Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule]
(ii) In ΔABE ΔACF [∵ Proved above]
See lessAB = AC [∵ CPCT]
Hence, ΔABC is an isosceles triangle.
ABC and DBC are two isosceles triangles on the same base BC (see Figure). Show that ∠ ABD = ∠ ACD.
In ΔABC, AB = AC [∵ Given] ∠ABC = ∠ACB ...(1) [∵ Angles opposite to equal sides are equal] In ΔDBC, DB = DC [∵ Given] ∠DBC = ∠DCB ...(2) [∵ Angles opposite to equal sides are equal] Adding equation (1) and (2), we get ∠ABC + ∠DBC = ∠ACB + ∠DCB ⇒ ∠ABD = ∠ACD.
In ΔABC,
See lessAB = AC [∵ Given]
∠ABC = ∠ACB …(1) [∵ Angles opposite to equal sides are equal]
In ΔDBC,
DB = DC [∵ Given]
∠DBC = ∠DCB …(2) [∵ Angles opposite to equal sides are equal]
Adding equation (1) and (2), we get
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒ ∠ABD = ∠ACD.