Question

Heya !!

➡ In an isosceles triangle, prove that the altitude from the vertex bisects the base.

➡ Answer if you know..

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Answer 1

hye

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here is ur answer 

1.CD is the altitude from vertex C to base AB.

2. Hence the triangles ADC and CDB are equal because they have two sides equal and one angle equal (SAS) then AD=DB .

3.it is 90degress

4.hence the altitude from the vertex   bisects the base.

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hope it helps u ....

Answer 2

hey dear



< riya 113 >



here is your answer




Solution



first of all I have to say figure of question in attachment you can see


< According to question >


Given - triangle PQR


Side PQ = PR


( PL is the bisector of an angle P)



To Prove - PLQ = PLR = 90 Degree

QL = LR


. ( 90 is the bisector of isosceles triangle)




In triangle PLQ and PLR

PQ = QR ( given)


PL = PL. ( common)


QPL = RPL.

( PL is the bisector of angle P)



hence PLQ. congruent to PLR

[ By SAS ( side angle side ) criteria ]


QL = LR

( by CPCT [ corresponding part of congruent triangle] )


and also PLQ = PLR ( by CPCT)



PLQ + PLR = 180 degree ( by linear pair)


2PLQ = 180


PLQ = 180 /2


PLQ = 90 degree



Hence PLQ = PLR =90 degree = QL = LR




hence it proved


< it prove that an isosceles triangle altitude of vertex bisect the base >




hope it helps


thank you