Pls. Solve this question in easy way of triangles
AP=BP [GIVEN]
AQ=QB [GIVEN]
AB=AB [COMMON]
BY S.S.S CRITERION OF CONGRUENCY WE CAN SAY THAT TRIANGLE APB=AQB
BY CPCT <PAB=<BAQ & <PBA=<ABQ
SO WE CAN SAY THAT
PLEASE LET ME KNOW THAT U UNDERSTOOD OR NOT....
IF U UNDERSTOOD, THEN PLS DONT FORGET TO LIKE MY ANSWER AND FOLLOW ME........
REGARDS..
BY-
TANISH
AQ=QB [GIVEN]
AB=AB [COMMON]
BY S.S.S CRITERION OF CONGRUENCY WE CAN SAY THAT TRIANGLE APB=AQB
BY CPCT <PAB=<BAQ & <PBA=<ABQ
SO WE CAN SAY THAT
- <PAB+<BAQ=<PAQ
- <PAB+<PAB=<PAQ [PROVED ABOVE THAT <PAB=<BAQ. SO WE CAN WRITE <PAB INSTEAD OF <BAQ AS BOTH ARE EQUAL.]
- <PBA+<ABQ=<PBQ[PROVED ABOVE THAT <PBA=<ABQ. SO WE CAN WRITE <PBA INSTEAD OF <ABQ AS BOTH ARE EQUAL.]
- SO THIS PROVES THAT AB IS THE BISECTOR OF THE GIVEN ANGLES WHICH YOU ASKED..
PLEASE LET ME KNOW THAT U UNDERSTOOD OR NOT....
IF U UNDERSTOOD, THEN PLS DONT FORGET TO LIKE MY ANSWER AND FOLLOW ME........
REGARDS..
BY-
TANISH