Question

15.
Let AABC be an acute angled A, If lines through A, B, C cut then opposite sides at D, E, F respectively such that AD, BE, CF make angle make angle
*% with BC, CA, AB respectively. Now AD and BE cut at Q, BE and CF cut at R. CF and AD out of at P. Then sin Arcos B+tan C
sin R+cos Puan​

Answer 1

$REF.Image.</p><p></p><p>Given :- AD⊥BC</p><p></p><p>AE=EC</p><p></p><p>FM=2</p><p></p><p>MN=1</p><p></p><p>NC=3</p><p></p><p>E,N or mid point of AC, CN </p><p></p><p>ΔADC;  EN∥AF</p><p></p><p>∠EDC=∠B.</p><p></p><p>△ADC;</p><p></p><p>EC=EA=DE</p><p></p><p>∠EDC=∠C</p><p></p><p>∠C=∠B</p><p></p><p>then AB=AC.</p><p></p><p>AD Bisect base BC. </p><p></p><p>Since AD divide CF such that</p><p></p><p>CM:MF=2:1</p><p></p><p>CF is median , point -F is mid point of AB.</p><p></p><p>EF∥BC</p><p></p><p>EF=2BC</p><p></p><p>∠EFC=∠FCB=∠ECF</p><p></p><p>EF=EC=AC/2</p><p></p><p>BC=AC=AB.</p><p></p><p>△ABC Becomes equilateral triangle</p><p></p><p>median = Angle Bisector = Altitude</p><p></p><p>CE=6</p><p></p><p>CE=23×AB</p><p></p><p>AB=312</p><p></p><p>Perimeter =123 unit </p><p></p><p>$

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