Question

8.ΔABC is isoceles with AB=AC, AD is the altitude from A to side BC,prove that *
1 point
(i) ΔADB≅ΔADC
(ii) ∠BAD=∠CAD
(iii)ΔACB≅ΔADC
(iv)ΔABC≅ΔADC
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Answer 1

Step-by-step explanation:

In tringle ABD & ACD

AB=AC (GIVEN)

ANGLE ADB=ANGLE ADC(BOTH ARE RIGHT ANGLES)

AD=AD(COMMON)

SO BOTH THE TRIANGLES ARE CONGRUENT

THEREFORE BD=CD(CPCT)

ANGLE BAD=ANGLE CAD(CPCT)

HENCE AD BISECTS BOTH BC AND ANGLE A

Answer 2

Answer:

Yes it is proved

Step-by-step explanation:

(i) We have AB = AC (Given) ∠BAD = ∠CAD (AD bisects ∠BAC) Therefore by SAS condition of congruence, ΔABD ≅ ΔACD (ii) We have used AB, AC; ∠BAD = ∠CAD; AD, DA. (iii) Now, ΔABD≅ΔACD Therefore by corresponding parts of congruent triangles BD = DC.