Question

in the adjoining figure AB= AC, CP || BA and AP is the bisector of angle CAD prove that angle PAC = angle BCA and ABCD is a parallelogram

This was the first question this is the second question



The diagonals AC and BD of a rhombus intersect each other at O prove that AB2+BC2+CD2+DA2=4 (OA2+OB2)

Answer 1

Solution 02

Diagonal's of Rhombus bisect at 90° and AO=OC & BO=OD

Apply pythogorous

AB²=AO²+BO²

BC²=BO²+OC²

CD²=OD²+OC²

AD²=OD²+AO²

AB²+BC²+CD²+AD²= [2AO²+2BO²+2CO²+2DO²]

AB²+BC²+CD²+AD²= 2[(AO+OC)²-2AO.OC+(BO+OD)²-2BO.OD]

AB²+BC²+CD²+AD²=2[(4OA²)-2OA²+(4OB²)-2OB²]

AB²+BC²+CD²+AD²=2[2OA²+2OB²]

AB²+BC²+CD²+AD²=2×2[OA²+OB²]

AB²+BC²+CD²+AD²=4[OA²+OB²]

Answer 2

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