Quadrilateral ABCD is a rhombus and AB is extended to P such that BP=AB . Prove that triangles DAB and CBP are congruent and diagonal BD is parallel to PC .
Proof : AB=AD=BC [Quadrilateral ABCD is a rhombus]
angle B=angle D
angle p=angle C [Both triangles ABD and PBC are isosceles triangles]
angle B+angle D = angle P + angle c
angle A=180-angle B-angle D
angle B=180-angle P-angle C
angle A=angle B
Hence , triangle DAB is congruent to triangle ABD
Since angle P = angle ABD [corresponding angle] , BD // PC
The straight line that passes through D , parallel to AC meets PC prodiced at R , Prove that .
I am not really sure bout this part. I cant see that which triangle is congruent to which in this case
THanks . This is the continuation of the question .
Originally Posted by Grandad
Given that LP=6 cm , PN=4 cm , and NM=5 cm , use similar triangles to calculate the length of LN .
I tried to find 2 triangles which are congruent . Triangle LPN and triangle LNM .
angle LPN = angle LNM
angle PLN=angle MLN
LN = LN
but one is SAA and the other one is ASA .They cant be congruent .