A triangle ABC has $∠$ B = $∠$ C.
Prove that :

(i) The perpendiculars from the mid-point of BC to AB and AC are equal.

(ii) The perpendiculars from B and C to the opposite sides are equal.

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Solution

Answer :

Given

B = C

i ) According to given information , we have our figure , As :

In triangle BDF and CDE

BD = CD ( Given )

Angle DBF = DCE ( Given )

And

Angle BFD = CED ( As we draw to perpendiculars on side AB and AC )

So,

tritrian BDF and CDE are congcongru ( By SAA rule )

Hence

DF = DE ( by CPCT ) ( Hence proved )

ii ) According to given information , we have our figure , As :

AnglA DBC = ECB ( Given )

Angle BDC = CEB ( As we draw to perpendiculars on side AB and AC )
And

BC = BC ( Common side )

So,

( By SAA rule )

Hence

BE = CD ( by CPCT ) ( Hence proved )

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A triangle ABC has ∠ B = ∠ C. Prove that : (i) The perpendiculars from the mid-point of BC to AB and AC are equal. (ii) The perpendiculars from B and C to the opposite sides are equal.

A triangle ABC has $∠$ B = $∠$ C.
Prove that :

(i) The perpendiculars from the mid-point of BC to AB and AC are equal.

(ii) The perpendiculars from B and C to the opposite sides are equal.