Ex 6.3, 7
In figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
ΔAEP ∼ ΔCDP
Given: Δ ABC
and , altitude AD and CE of triangle
intersects each other at the point P.
To Prove : ΔAEP ∼ ΔCDP
Proof:
In ΔAEP and ΔCDP
∠AEP = ∠CDP
∠APE = ∠CPD
Hence, ΔAEP ∼ ΔCDP
Hence proved
Ex 6.3,7
In figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(ii) ΔABD ∼ ΔCBE
In Δ ABD & Δ CBE
∠ABD = ∠CBE
∠ADB = ∠CEB
By using AA similarity
ΔABD ∼ ΔCBE
Ex 6.3, 7
In figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(iii) ΔAEP ∼ ΔADB
In ΔAEP and ΔADB
∠PAE = ∠DAB
∠AEP = ∠ADB
By using AA similarity
So, ΔAEP ∼ ΔADB
Ex 6.3, 7
In figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(iv) ΔPDC ∼ ΔBEC
In ΔBEC and ΔPDC
∠BCE = ∠PCD
∠CEB = ∠CDP
By using AA similarity
So, ΔBEC ∼ ΔPDC
Hence proved

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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