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Obtuse Angled Triangle

33. In triang...

Question

33. In triangle ABC,segAB = SegA C & P is any points on AC.Through C, a line is drawn to intersect BP produced in Q such that ANGLE ABQ=ANGLE ACQ .Prove that ANGLE AQC=90^° +1/2 ANGLE BAC.

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△

∠

∠

∠ ACQ = 90 ° + \frac{1}{2} ∠ BAC

∠ B

∠ C

△ A B C

∠ B P C = 90 + \frac{1}{2} ∠ B A C

A B C

A B = A C

P

Δ A B C

∠ A B P = ∠ A C P

B P = C P

A P

∠ B A C

In triangle ABC; angle ABC = $90^{0}$ and P is a point on AC such that $∠$ PBC = $∠$ PCB. Show that : BP = CP.

∆

∠

∠

∠

- \frac{1}{2}

∠

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