Question

ΔABC is isosceles with AB=AC, AD is the altitude from A to side BC, prove that ΔADB ≅ΔADC. Draw a well-labeled figure.

Answer 1

In the above triangles ADB and ADC:

Angle ADB = Angle ADC ( AD is an altitude : given)

Angle ABD = Angle ACD ( Two angles of an isosceles triangle are equal)

Therefore, By A-A test of similarity.... Triangle ADB is similar to Triangle ADC.

Hope this helps.

Answer 2

Answer:

Step-by-step explanation:

Given: AB = AC

Let's construct an isosceles triangle ABC in which AB = AC as shown below.

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects ∠A

(i) In ΔBAD and ΔCAD,

∠ADB = ∠ADC (Each 90° as AD is an altitude)

AB = AC (Given)

AD = AD (Common)

∴ ΔBAD ≅ ΔCAD (By RHS Congruence rule)

∴ BD = CD (By CPCT)

Hence, AD bisects BC.

(ii) Since, ΔBAD ≅ ΔCAD

By CPCT, ∠BAD = ∠CAD

Hence, AD bisects ∠A.