Math, asked by js384202, 1 year ago

Prove that the perpendicular drawn to the base of an isosceles triangle from the vertex bisect the base??


js384202: sove
js384202: solve kar do
js384202: any other answer

Answers

Answered by shashankdangwal123
21

Answer:

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.

hope it will help you



js384202: copy cha solve kar
shashankdangwal123: kya
js384202: isoseles triangle were want you not drawn
js384202: in copy solve this question in short
shashankdangwal123: what the F**k
js384202: your answer was to long i cannot understand
shashankdangwal123: so what can i do
js384202: your answer is wrong
js384202: okkk
deepasreebiju: Lol
Answered by ashpatel38
38
Hope it helps.... plsss make it the brainliest
Attachments:

ashpatel38: as they are common sides of two triangles
ashpatel38: CPCT = corresponding part of congruent triangle
js384202: thanks thanks
js384202: i marked you
ashpatel38: above it I have written both the triangles are congruent
ashpatel38: welcome
js384202: can you solve my 1 more question
ashpatel38: which subject
js384202: maths
ashpatel38: go on
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