ABCD is a Rhombus. DPR and CBR are straight lines. Prove that: DP.CR=DC.PR

ABCD is a rhombus. DPR and CBR are straight lines.
Prove that:
DP.CR=DC.PR

Solution:

DP.CR=DC.PR
Given ABCD is rhombus .
DPR and CBR are straight lines.
AD||CR
we need to Prove : DP.CR=DC.PR
In &#x2206 ADP and &#x2206 PCR
We have :
&#x2220 APD = &#x2220 CPR
&#x2220 ADP = &#x2220 PRC
&#x2220 DAP = &#x2220 PCR
&#x2206 ADP and &#x2206 PCR are similar triangle .
we can write
AD/DP=CR/PR
Or AD.PR = DP.CR
DC .PR = DP.CR Proved

See also  OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB of &#x2206 ABC |Prove that AR2 +BP2 +CQ2 = AQ 2 + CP 2+BR2

Comments

2 responses to “ABCD is a Rhombus. DPR and CBR are straight lines. Prove that: DP.CR=DC.PR”

  1. For two similar triangles [ADP and PCR] which angles are equal. The ratio of sides of one angle can be equal to the ratio of sides of other triangle . Please read about similar triangles , you can get this property. Hope I am able to clarify your query.

  2. How is AD/DP?

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