Prove that the perpendicular drawn to the base of an isosceles triangle from the vertex bisect the base??
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Consider PQR is an isosceles triangle such that PQ = PR and Pl is the bisector of ∠ P.
To prove : ∠PLQ = ∠PLR = 90°
and QL = LX
In ΔPLQ and ΔPLR
PQ = PR (given)
PL = PL (common)
∠QPL = ∠RPL ( PL is the bisector of ∠P)
ΔPLQ = ΔPLR ( SAS congruence criterion)
QL = LR (by cpct)
and ∠PLQ + ∠PLR = 180° ( linear pair)
2∠PLQ = 180°
∠PLQ = 180° / 2 = 90°
∴ ∠PLQ = ∠PLR = 90°
Thus, ∠PLQ = ∠PLR = 90° and QL = LR.
Hence, the bisector of the verticle angle an isosceles triangle bisects the base at right angle.
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