Question

Consider two triangular parks ABC and EFG such that a path from one corner of the park bisects the other side of the park as shown in the figure. If the dimensions of the park ABC are AB = 4 m, BC = 6 m and AC = 3 m and that of park EFG are EF = 2 m, EG = 3 m and FG = 1.5 m.

Which of the following statements are true?

ΔADC∼ΔFHG

CDGH=ABFE

ΔCDB∼GHE

CD = GH

Solution

The correct options are

**A**

**B**

**C**

ΔADC∼ΔFHG

CDGH=ABFE

ΔCDB∼GHE

In ΔABC and ΔEFG

AB = 4 m

BC = 6 m

AC = 3 m

EF = 2 m

FG = 1.5 m

and EG = 3 m

CD and GH are medians to sides AB and EF, respectively.

Now in ΔABC and ΔEFG,

ABBC=EFEG

and BCAC=EGFG

∴ΔABC∼ΔFEG…(i) [by SSS similarity]

⇒∠A=∠F,∠B=∠E & ∠C=∠G…(ii)

Now, ABFE=ACFG=BCEG⇒2AD2FH=ACFG=BCEG⇒ADFH=ACFG=BCEG…(iii)

(a) In ΔADC and ΔFHG, we have

ADFH=ACFG

and ∠A=∠F (from Eq. (iii))

∴ΔADC∼ΔFHG [by SAS similarity]

(b) We have,ΔADC∼ΔFHG⇒DCHG=ADFH⇒CDGH=2AD2FH⇒CDGH=ABFE[∵AB=2AD,FE=2FH]

(c) Now, ABFE=ACFG=BCEG…(iv) [from Eq. (i)]

Also, CDGH=ABFE

⇒CDGH=BCEG [from Eq. (iv)]

∴ΔCDB∼ΔGHE [by SSS similarity]

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