Triangles Proof Problem!

Asked by RACHIT_MAHAAN | 30th Dec, 2009, 01:07: PM

Expert Answer:

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.

so

PD/PA=PB/PC.. corr sides of similar triangles  are proportional.

so

PA.PB=PC.PD

Note : the result must be as shown in bold letters.

 

Answered by  | 30th Dec, 2009, 06:52: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.