Question

In the figure,given below, the medians BD and CE of a triangle ABC meet at G. Prove that :
i)ΔEGD ~ ΔCGB, and
ii)BG=2GD from (i) above.

Answer 1

$\huge\underline\mathfrak\orange{Answer}$

(i) Since, BD and CE are medians. AD = DC AE = BE Hence, by converse of Basic Proportionality theorem, ED || BC In ΔEGD and ΔCGB, ∠DEG = ∠GCB (alternate angles) ∠EGD = ∠BGC (Vertically opposite angles) ΔEGD ~ ΔCGB (AA similarity) (ii) since, ΔEGD ~ ΔCGB in-the-figure-given-below-the-medians-bd-and-ce-of-a-triangle-abc-meet-at-g-prove-that .

Answer 2

Step-by-step explanation:

(i)

Since, BD and CE are medians.

AD = DC

AE = BE

Hence, by converse of Basic Proportionality theorem,

=> ED || BC

In ∆EGD and ∆CGB,

=> ∠DEG = ∠GCB (alternate angles)

=> ∠EGD = ∠BGC (Vertically opposite angles)

∴ ∆EGD ~ ∆CGB

(ii)

Since, ∆EGD ~ ∆CGB

GD/GB = ED/BC   ----- (1)

In ∆AED and ∆ABC,

=> ∠AED = ∠ABC (Corresponding angles)

=> ∠EAD = ∠BAC (Common)

=> ∆EAD ~ ∆BAC (AA similarity)

(ED/BC) = (AE/AB) = 1/2

=> ED/BC = 1/2

From (1),

(GD/DB) = 1/2

∴ GB = 2GD

Hope it helps!