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 ΔADC∼ΔFHG   B CDGH=ABFE C 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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