Question

plz solve question no 30

Answer 1

∆ABC is an isosceles triangle,

we know that, isosceles triangle have two sides equal,

AB = AC
AP+BP = AQ+CQ
AQ+BP = AQ+CQ ( AQ=AP ,it is given)
BP = CQ

angle ABC=angle ACQ (two sides are equal,then angle also equal)
therefore,
angle PBC=angle QCB

In ∆PBC and ∆QCB,
BP = CQ (proved above)
angle PBC = angle QCB (we proved above)
BC=BC (common)

∆PBC ~= ∆QCB ( By SAS congurency)

CP = QB ( by cpct)

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