similarity question
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if u notice ∆abc, u get, <ABC=90 & ab2+bc2=ac2
if u notice &rotate ∆pqr, u will get, <pqr=90 and also qr2+pq2=pr2
as, 2^2+(1.5)^2=(2.5)^2
so they r similar for right angles
<ABC=<pqr
so, ∆abc= =∆pqr for angle angle symmetry
if u notice &rotate ∆pqr, u will get, <pqr=90 and also qr2+pq2=pr2
as, 2^2+(1.5)^2=(2.5)^2
so they r similar for right angles
<ABC=<pqr
so, ∆abc= =∆pqr for angle angle symmetry
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