In the given figure (not drawn to scale), BC is parallel to EF, CD is parallel to FG, AE : EB = 2 : 3, ∠BAD = 70°, ∠ACB = 105°, ∠ADC = 40° and AC is bisector of ∠BAD.
(a) Prove Δ AEF ~ Δ AGF
(b) Find :
(i) AG : AD
(ii) area of Δ ACB: area Δ ACD
(iii) area of quadrilateral ABCD: area of Δ ACB.

Question

In the given figure (not drawn to scale), BC is parallel to EF, CD is parallel to FG, AE : EB = 2 : 3, ∠BAD = 70°, ∠ACB = 105°, ∠ADC = 40° and AC is bisector of ∠BAD.
(a) Prove Δ AEF ~ Δ AGF
(b) Find :
(i) AG : AD
(ii) area of Δ ACB: area Δ ACD
(iii) area of quadrilateral ABCD: area of Δ ACB.

KnowledgeBoat · Accepted Answer

(a) In Δ AFE, ⇒ ∠AFE = ∠ACB = 105° (Corresponding angle are equal) ⇒ ∠EAF = = 35° (AC is the bisector of ∠BAD) By angle sum property of triangle, ⇒ ∠AFE + ∠EAF + ∠AEF = 180° ⇒ 105° + 35° + ∠AEF = 180° ⇒ 140° + ∠AEF = 180° ⇒ ∠AEF = 180° - 140° = 40°. ∠AGF = ∠ADC = 40° (Corresponding angles are equal) In Δ AGF, ⇒ ∠GAF = = 35° (AC is the bisector of ∠BAD) In Δ AEF and Δ AGF, ⇒ ∠EAF = ∠GAF = 35° (AC being the bisector) ⇒ ∠AEF = ∠AGF = 40° (Proved above) ∴ Δ AEF ~ Δ AGF (By A.A. axiom) Hence, proved that Δ AEF ~ Δ AGF. (b) In Δ ACD, ⇒ ∠CAD = 35° ⇒ ∠ADC = 40° By angle sum property of triangle, ⇒ ∠CAD + ∠ADC + ∠DCA = 180° ⇒ 35° + 40° + ∠DCA = 180° ⇒ ∠DCA + 75° = 180° ⇒ ∠DCA = 180° - 75° = 105°. In Δ ACD and Δ ACB, ⇒ ∠ACB = ∠ACD = 105° (Proved above) ⇒ ∠BAC = ∠DAC = 35° (Proved above) ∴ Δ ACD ~ Δ ACB (By A.A. axiom) (i) Given, ⇒ AE : EB = 2 : 3 Let AE = 2x and EB = 3x. ∴ AB = AE + EB = 2x + 3x = 5x. We know that, Ratio of corresponding sides are proportional. Hence, AG : AD = 2 : 5. (ii) In △ ABC and △ ADC, ⇒ ∠BAC = ∠DAC (Both equal to 35°) ⇒ ∠ACB = ∠ACD (Both equal to 105°) ⇒ ∠ABC = ∠ADC (Both equal to 40°) ∴ △ ABC and △ ADC are congruent. We know that, Area of congruent triangles are equal. Let area of △ ABC and area of △ ADC = x. ∴ Area of △ ABC : Area of △ ADC = x : x = 1 : 1. Hence, area of △ ABC : area of △ ADC = 1 : 1. (iii) From figure, Area of quadrilateral ABCD = Area of △ ABC + Area of △ ADC = x + x = 2x. ∴ Area of quadrilateral ABCD : Area of △ ACB = 2x : x = 2 : 1. Hence, area of quadrilateral ABCD : area of △ ACB = 2 : 1.

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