Triangles Proof Problem!
Asked by RACHIT_MAHAAN | 30th Dec, 2009, 01:07: PM
the quadrilateral ABDC is a cyclic quadrilateral as one pair of opp angles is supplementary.
so PDB being the exterior angle of the cyclic quadrilateral is equal to the interior opp angle i.e. angel BAC.
now, consider triangles PDB and PAC,
angle P = angle P ... common
angle PDB=angle PAC.. proved before since PAC is the same angle BAC
so, the two triangles are similar by AA rule.
PD/PA=PB/PC.. corr sides of similar triangles are proportional.
Note : the result must be as shown in bold letters.
Answered by | 30th Dec, 2009, 06:52: PM
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