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Asked by Nagarunja Last Modified
Answer 
Sadika
To prove that the lengths of altitudes drawn to equal sides of an isosceles triangle are also equal:
(i) Given that in triangle PQR, PR = PQ (isosceles triangle), and altitudes QT and RS are drawn to equal sides PQ and PR respectively.
We'll prove that angle TRQ is equal to angle SQR.
Given: PR = PQ (isosceles triangle)
Prove: ∠TRQ = ∠SQR
Proof:
Since QT and RS are altitudes, they are perpendicular to their respective bases.
∠QTR = 90° and ∠SRP = 90°
Since triangle PQR is isosceles, PR = PQ, so base angles ∠PQR = ∠PRQ.
In △QTR and △SRP:
∠TRQ = ∠PQR (Alternate Interior Angles)
∠SQR = ∠PRQ (Alternate Interior Angles)
Since ∠PQR = ∠PRQ (base angles of an isosceles triangle are equal),
∠TRQ = ∠SQR
Thus, we have proved that angle TRQ is equal to angle SQR.
(ii) If ∠TRQ = 30°, then using the fact that ∠TRQ = ∠SQR (as proved above), ∠SQR = 30°.
Since the sum of angles in a triangle is 180°, and we know that ∠SQR = ∠SQR = 30°, we can find the measure of the base angles:
∠PQR = ∠PRQ = (180° - 30° - 30°) / 2 = (180° - 60°) / 2 = 120° / 2 = 60°
So, the base angles of triangle PQR are both 60°.
(iii) Since we found that the base angles of triangle PQR are both 60°, and they are equal, but not all sides are equal, triangle PQR is an isosceles triangle, but not equilateral.
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