Question

In triangle abc, the median ad is perpendicular to BC. Prove that ABC is a isosceles triangle .

Answer 1

Answer:Consider triangles ABD and ACD.

As AD is a median of ABC, D is the midpoint of BC.  So BD = CD.

The side AD is common to both triangles.

If AD is perpendicular to BC, then ∠ADB = 90° = ∠ADC.

So by the SAS rule, triangles ABC and ACD are congruent.

Therefore AB = AC.

It follows that ABC is isosceles

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Step-by-step explanation:

Answer 2

Given: line AD is perpendicular to BC

so, ∠ADC= ∠ADB = 90°⇒(1)

And line AD bisects line BC

⇒ BD = CD ⇒(2)

To prove: ΔABC is isosceles, i.e. AB = AC

Proof: In ΔABC and ΔACD,

AD = AD (common)

∠ADB = ∠ADC (from (1))

BD = CD (from(2))

∴ΔADC ≅ ΔADB (By SAS Congruence Criterion)

⇒AB = AC (C.P.C.T)

Therefore, ABC is an isosceles triangle in which AB = AC

Hence proved