If tanA =tanP ,than prove that …

Consider two right triangles XAY and WBZ such that tan A = tan B

We have,
tan A = {tex}\frac { X Y } { A Y }{/tex} and tan B = {tex}\frac { WZ } { B Z }{/tex}
Since, tan A = tan B
{tex}\Rightarrow \frac { X Y } { A Y } = \frac { W Z } { B Z }{/tex}
{tex}\Rightarrow \frac { X Y } { W Z } = \frac { A Y } { B Z } = k{/tex}(say)......(i)
{tex}\Rightarrow X Y = k \times W Z \text { and } A Y = k \times B Z{/tex}.......(ii)
Using pythagoras theorem in triangles XAY and WBZ, we have
XA² = XY² + AY² and WB² = WZ² + BZ²
{tex}\Rightarrow{/tex} XA² = k²WZ² + k²BZ² and WB² = WZ² + BZ²
{tex}\Rightarrow{/tex} XA² = k²(WZ² + BZ²) and WB² = WZ² + BZ²
{tex}\Rightarrow\frac { X A ^ { 2 } } { W B ^ { 2 } } = \frac { k ^ { 2 } \left( W Z ^ { 2 } + B Z ^ { 2 } \right) } { \left( W Z ^ { 2 } + B Z ^ { 2 } \right) } = k ^ { 2 }{/tex}
{tex}\Rightarrow \frac { X A } { W B } = k{/tex}...........(iii)
From (i), (ii) and (iii), we get
{tex}\frac { X Y } { W Z } = \frac { A Y } { B Z } = \frac { X A } { W B }{/tex}
{tex}\Rightarrow \Delta A Y X \sim \Delta B Z W{/tex}
{tex}\Rightarrow \angle A = \angle B{/tex}