Advertisements
Advertisements
प्रश्न
In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.
Advertisements
उत्तर

In ΔABC
E and F are midpoints of AB and AC
∴ EF || FE, `1/2` BC =FE [ ∴ By mid-point theorem]
In ΔABP
F is the midpoint of AB and FQ || BP [ ∵ EF || BC ]
∴ Q is the midpoint of AP [By converse of midpoint theorem]
Hence, AQ = QP
APPEARS IN
संबंधित प्रश्न
ABCD is a rhombus. EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced, meet at right angles.
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.

ABC is a triang D is a point on AB such that AD = `1/4` AB and E is a point on AC such that AE = `1/4` AC. Prove that DE = `1/4` BC.
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
L and M are the mid-point of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.
In Δ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.
In ΔABC, D is the mid-point of AB and E is the mid-point of BC.
Calculate:
(i) DE, if AC = 8.6 cm
(ii) ∠DEB, if ∠ACB = 72°
In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find DE, if AB = 8 cm
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: ∠EFG = 90°
In ΔABC, D and E are the midpoints of the sides AB and AC respectively. F is any point on the side BC. If DE intersects AF at P show that DP = PE.
