Check sibling questions

Ex 6.3, 7 - Altitudes AD and CE of ABC intersect each other - AA Similarity

Ex 6.3, 7 - Chapter 6 Class 10 Triangles - Part 2

Ex 6.3, 7 - Chapter 6 Class 10 Triangles - Part 3

Ex 6.3, 7 - Chapter 6 Class 10 Triangles - Part 4


Transcript

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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.