Given that B is a point on line AC, M is the midpoint of line AB, n is the midpoint of line AC, P is the midpoint of Na, q is the midpoint of Ma, find Mn: PQ Because PQ = ap-aq = an ﹣ 2-AM ﹣ 2 = (an-am) ﹣ 2 = Mn ﹣ 2 So Mn △ 2 = PQ Mn = 2pq So Mn: PQ = 2:1

Given that B is a point on line AC, M is the midpoint of line AB, n is the midpoint of line AC, P is the midpoint of Na, q is the midpoint of Ma, find Mn: PQ Because PQ = ap-aq = an ﹣ 2-AM ﹣ 2 = (an-am) ﹣ 2 = Mn ﹣ 2 So Mn △ 2 = PQ Mn = 2pq So Mn: PQ = 2:1

Solution ∵ m is the midpoint of ab ∵ am = BM = AB / 2 ∵ q is the midpoint of Ma ∵ AQ = QM = am / 2 = AB / 4 ∵ n is the midpoint of AC ∵ an = CN = AC / 2 ∵ P is the midpoint of Na ∵ AP = NP = an / 2 = AC / 4 ∵ Mn = an-am = AC / 2-AB / 2 = (ac-ab) / 2pq = ap-aq = AC / 4-ab / 4 = (ac-ab) / 4 ∵ Mn: PQ =

As shown in the figure, ab ∥ CD, ad ∥ CE, F and G are the midpoint of AC and FD respectively. The straight line passing through G intersects AB, ad, CD and CE successively at points m, N, P and Q. the verification is: Mn + PQ = 2pn

It is proved that if the intersection of Ba and EC is O, then the quadrilateral OADC is parallelogram, ∵ f is the midpoint of AC, the extension of ∵ DF must pass through O, and dgog = 13. ∵ ab ∥ CD, ∵ mnpn = andn. ∵ ad ∥ CE, ∵ pqpn = cqdn. ∵ mnpn + pqpn = andn + cqdn = an + cqdn. And ∵ dnoq = dgog = 13

As shown in the figure, in the acute angle △ ABC, ad and CE are the heights of BC and AB, respectively. AD and CE intersect at F, the midpoint of BF is p, and the midpoint of AC is Q, connecting PQ and de. (1) prove that the straight line PQ is the vertical bisector of the line segment de; (2) if △ ABC is an obtuse triangle, ∠ BAC > 90 °, then the above conclusion is true? Please rewrite the original title according to obtuse triangle, draw corresponding figure and give necessary explanation

(1) Because CE ⊥ AB, P is the midpoint of BF, so △ bef is a right triangle, and PE is the middle line of RT △ bef hypotenuse, so PE = 12bf. Because ad ⊥ BC, so △ BDF is a right triangle, and PD is the middle line of RT △ BDF hypotenuse, so PD = 12bf = PE, so

In rectangular ABCD, points E and F are on the edge of AD, CE and CF intersect with BD at points m and N, AE = EF = FD = 4cm, ab = 16cm, respectively

It can be verified that triangle EMD is similar to triangle MBC
So MD / MB = ed / BC = 8 / 12 = 2 / 3
Because BD = radical (16 * 16 + 12 * 12) = 20
So MD = 8
Similarly, it can be verified that the triangle fnd is similar to the triangle BNC
So nd = 5
So Mn = md-nd = 3