CD and GH are respectively the bisectors of Δ ACB and Δ EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that: (i) CD/GH = AC/FG (ii) Δ DCB ~ Δ HGE (iii) Δ DCA ~ Δ HGF

Question

KnowledgeBoat · Accepted Answer

(i) Given, ∆ ABC ~ ∆ FEG We know that, If two triangles are similar, then their corresponding angles are congruent. ⇒ ∠ACB = ∠FGE ⇒ ⇒ ∠ACD = ∠FGH (CD and GH are bisectors of ∠C and ∠G respectively) ..............(1) In ∆ ADC and ∆ FHG, ⇒ ∠DAC = ∠HFG [∵ ∆ABC ~ ∆FEG] ⇒ ∠ACD = ∠FGH [From equation (1)] ∴ ∆ ADC ~ ∆ FHG (By A.A. axiom) We know that, Ratio of corresponding sides in similar triangle are proportional. ∴ . Hence, proved that . (ii) In ∆ DCB and ∆ HGE, ⇒ ∠DBC = ∠HEG [∵ ∆ ABC ~ ∆ FEG] ⇒ ∠DCB = ∠HGE [∵ ] ∴ ∆ DCB ~ ∆ HGE (By A.A. axiom) Hence, proved that ∆ DCB ~ ∆ HGE. (iii) In ∆ DCA and ∆ HGF, ⇒ ∠DAC = ∠HFG [∆ ABC ~ ∆ FEG] ⇒ ∠ACD = ∠FGH [From (1)] ∴ ∆ DCA ~ ∆ HGF (By A.A. axiom) Hence, proved that ∆ DCA ~ ∆ HGF.

Mathematics

CD and GH are respectively the bisectors of Δ ACB and Δ EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that: (i) CD/GH = AC/FG (ii) Δ DCB ~ Δ HGE (iii) Δ DCA ~ Δ HGF

Triangles

Related Questions

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that △ ABE ~ △ CFB.

In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that △ ABD ~ △ ECF.

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of △ PQR. Show that △ ABC ~ △ PQR.